Paths Have Prescribed Endpoint Values
A path is not just a continuous curve in a space: its values at the two ends of its parameter interval are part of the data. Once points \(x\) and \(y\) are prescribed, a path from \(x\) to \(y\) must start at \(x\) and finish at \(y\). A formula that traces a plausible curve but fails either endpoint condition does not give a path between those prescribed points.
Throughout, \(I=[0,1]\) has the subspace topology inherited from the standard topology on \(\mathbb{R}\), and \((X,\mathcal{T})\) is a topological space. A map \(\gamma:I\to X\) is continuous in the usual topological sense: for every open set \(U\) in \(X\), the preimage \(\gamma^{-1}(U)\) is open in \(I\).
The endpoint equations are separate from continuity: to verify a proposed path, check both that its values lie in \(X\) and that the map is continuous, then evaluate it at \(0\) and \(1\). Its image must contain both endpoints, since \(\gamma(0)=x\) and \(\gamma(1)=y\). The endpoint conditions do not, by themselves, specify the other values of the path.
Constant Paths and Their Endpoint Criterion
Proof. Fix \(x\in X\). For any open set \(U\in\mathcal{T}\), the preimage \(c_x^{-1}(U)\) is \(I\) if \(x\in U\), and is \(\varnothing\) if \(x\notin U\). Both \(I\) and \(\varnothing\) are open in \(I\), so \(c_x\) is continuous. Its endpoint values are \(c_x(0)=x\) and \(c_x(1)=x\), making it a path from \(x\) to \(x\). If a constant map has value \(z\), then both endpoint values equal \(z\). Therefore, if it is a path from \(x\) to \(y\), then \(x=z=y\). Conversely, when \(x=y\), the map \(c_x\) just constructed is a constant path from \(x\) to \(y\). \(\square\)
This criterion concerns constant paths, not all paths. A path whose initial and terminal points agree need not be constant: it may leave that point and later return. Conversely, a path from distinct \(x\) and \(y\), if one exists, cannot be constant.
Worked Example: Prescribed Endpoints in an Interval
Give a path in \(I=[0,1]\), with its subspace topology from \(\mathbb{R}\), from \(a=\frac14\) to \(b=\frac34\). Check the endpoints and that the values remain in \(I\).
Consider \(\gamma:I\to I\) given by \(\gamma(t)=\frac14+\frac12t\). For every \(t\in I\), \(0\leq t\leq1\), so \(\frac14\leq\frac12t+\frac14\leq\frac34\). Thus \(\gamma(t)\in I\). The formula is affine, hence continuous as a real-valued function; because its image lies in \(I\), it is also continuous as a map into the subspace \(I\).
At the initial parameter, \(\gamma(0)=\frac14+\frac12(0)=\frac14=a\). At the terminal parameter, \(\gamma(1)=\frac14+\frac12(1)=\frac14+\frac12=\frac34=b\). As an additional value check, \(\gamma(\frac12)=\frac14+\frac12\cdot\frac12=\frac12\), which is in \(I\). Therefore \(\gamma\) is a path from \(a\) to \(b\).
Check: the endpoint values obtained by substitution are exactly the prescribed values, and the bound \(\frac14\leq\gamma(t)\leq\frac34\) holds for every permitted parameter \(t\). The map stays in the required space, not merely in \(\mathbb{R}\).
There can be more than one path with the same endpoints. For example, on this same interval, the function \(\eta(t)=\frac14+\frac12t+\frac18\sin(2\pi t)\) has \(\eta(0)=\frac14\) and \(\eta(1)=\frac34\). It stays in \(I\): since \(-1\leq\sin(2\pi t)\leq1\), we have \(\eta(t)\geq\frac18+\frac12t\geq\frac18\) and \(\eta(t)\leq\frac38+\frac12t\leq\frac78\). It is continuous, and at \(t=\frac12\) its value is \(\frac12\), whereas \(\gamma(\frac12)=\frac12\) as well; at \(t=\frac14\), however, \(\eta(\frac14)=\frac14+\frac18+\frac18=\frac12\), while \(\gamma(\frac14)=\frac38\). Thus the two paths have the same endpoints but are different maps.
Endpoint Checks on the Circle
Let \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) have the subspace topology inherited from the Euclidean topology on \(\mathbb{R}^2\). A parametrized pair of coordinates gives a path in \(S^1\) only if it stays on the circle, is continuous, and has the required endpoint values.
Worked Example: A Quarter-Circle Path
Construct a path from \((1,0)\) to \((0,1)\) in \(S^1\). Use \(\gamma(t)=(\cos(\frac{\pi t}{2}),\sin(\frac{\pi t}{2}))\) for \(t\in I\).
The sine and cosine functions are continuous, so the coordinate map into \(\mathbb{R}^2\) is continuous. For every \(t\), the identity \(\cos^2(\frac{\pi t}{2})+\sin^2(\frac{\pi t}{2})=1\) shows that \(\gamma(t)\in S^1\). Consequently the same map, with codomain restricted to the subspace \(S^1\), is continuous.
At \(t=0\), the angle is \(0\), so \(\cos(0)=1\) and \(\sin(0)=0\); hence \(\gamma(0)=(1,0)\). At \(t=1\), the angle is \(\frac{\pi}{2}\), so \(\cos(\frac{\pi}{2})=0\) and \(\sin(\frac{\pi}{2})=1\); hence \(\gamma(1)=(0,1)\).
Check: the initial and terminal values are the stated points, and both lie on \(S^1\), since \(1^2+0^2=1\) and \(0^2+1^2=1\). Thus the formula defines the required path.
Worked Example: Equal Endpoints Do Not Force a Constant Path
In \(S^1\), compare the constant path at \((1,0)\) with \(\lambda(t)=(\cos(2\pi t),\sin(2\pi t))\).
The coordinate functions are continuous, and \(\cos^2(2\pi t)+\sin^2(2\pi t)=1\) for every \(t\), so \(\lambda\) is a continuous map into \(S^1\). At \(t=0\), its angle is \(0\), giving \(\lambda(0)=(1,0)\). At \(t=1\), its angle is \(2\pi\), giving \(\lambda(1)=(1,0)\), because \(\cos(2\pi)=1\) and \(\sin(2\pi)=0\). These are the same endpoint values as the constant path \(c_{(1,0)}\).
But at \(t=\frac12\), the angle is \(\pi\), so \(\lambda(\frac12)=(\cos\pi,\sin\pi)=(-1,0)\). This differs from \((1,0)\), the value of the constant path at every parameter. Therefore \(\lambda\) is not constant, even though its two endpoints agree.
Check: \((-1,0)\in S^1\) because \((-1)^2+0^2=1\). The example satisfies both endpoint conditions while taking a different value between them.
Paths into a Discrete Space
A space \(X\) is discrete if every subset of \(X\) is open. In such a space, the endpoint requirements have a strong consequence: no continuous path can move between distinct points. The reason is that the interval \(I\) is connected, whereas a continuous map cannot send a connected space onto a set that splits into two nonempty disjoint open parts. Here is a direct proof for a discrete target.
Proof. Suppose, for a contradiction, that \(\gamma\) is not constant. Choose \(t_0,t_1\in I\) with \(\gamma(t_0)\ne\gamma(t_1)\), and put \(p=\gamma(t_0)\). The singleton \(\{p\}\) is open in \(X\), and its complement \(X\setminus\{p\}\) is open because \(X\) is discrete. Continuity therefore makes \(A=\gamma^{-1}(\{p\})\) and \(I\setminus A=\gamma^{-1}(X\setminus\{p\})\) open in \(I\). Both are nonempty: \(t_0\in A\), while \(t_1\in I\setminus A\). They are disjoint and their union is \(I\). This is a separation of \(I\), contradicting the connectedness of the interval. Hence \(\gamma\) must be constant. \(\square\)
Worked Example: Prescribed Endpoints in a Discrete Space
Let \(X=\{p,q,r\}\) have the discrete topology. Determine whether there is a path from \(p\) to \(q\), and give a path from \(r\) to \(r\).
If there were a path \(\gamma:I\to X\) from \(p\) to \(q\), the theorem would force \(\gamma\) to be constant. A constant map has equal initial and terminal values, but the required values \(p\) and \(q\) are distinct. Therefore no such path exists. For endpoints \(r,r\), define \(c_r(t)=r\) for every \(t\in I\). The constant-path proposition shows that \(c_r\) is continuous, and substitution gives \(c_r(0)=r=c_r(1)\).
Check: the impossibility for \(p,q\) uses \(p\ne q\); the existence for \(r,r\) uses equal endpoints. These conclusions apply to a discrete space of any size, including an infinite discrete space, by the theorem.
A Common Endpoint Mistake
Common pitfall. A proposed formula can have the right starting value and still fail to be a path between the stated endpoints. For instance, the constant map \(c_{\frac14}(t)=\frac14\) has \(c_{\frac14}(0)=\frac14\), but substituting the terminal parameter gives \(c_{\frac14}(1)=\frac14\), not \(\frac34\). It is a path from \(\frac14\) to \(\frac14\), not from \(\frac14\) to \(\frac34\). The correction is to check both substitutions: for \(\gamma(t)=\frac14+\frac12t\), they give \(\gamma(0)=\frac14\) and \(\gamma(1)=\frac34\). Even after these checks, continuity and membership in the target space must still be verified.
The endpoint test is short but decisive: write down the proposed domain and codomain, verify continuity and that the map takes values in the codomain, then compute the values at \(0\) and \(1\). In a discrete target, the theorem above can rule out paths with distinct endpoints before any formula is attempted. In other spaces, the endpoint equations constrain a path without generally determining it uniquely.
Check Your Understanding
For each question, distinguish the endpoint calculation from the continuity or target-space check.
- For \(a,b\in\mathbb{R}\), evaluate the endpoints of \(\gamma(t)=(1-t)a+tb\). What additional condition ensures that this formula maps \(I\) into a specified subset of \(\mathbb{R}\)?
- Can a constant path have distinct initial and terminal points? Justify your answer from its endpoint values.
- In \(S^1\), evaluate \((\cos(\pi t),\sin(\pi t))\) at \(t=0\) and \(t=1\). Which points are its endpoints?
- Why can there be no path between distinct points of a discrete space?
- Give an example of a nonconstant path whose initial and terminal points agree, and verify its values at \(0\), \(\frac12\), and \(1\).